In: Chemistry

# Calculate the pH for each of the following cases in the titration of 50.0 mL of...

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.150 M CH3COOH(aq) with 0.150 M KOH(aq). The ionization constant for HClO is Ka = 1.8 x 10-5. Just give the number to 2 decimal places.

A) Before addition of any KOH: pH =

B) After addition of 25 mL of KOH: pH =

C) After addition of 50 mL of KOH: pH =

D) After addition of 60 mL of KOH: pH

## Solutions

##### Expert Solution

This is a titration, so we will have different pH at different Volumes:

First of all, volumes will change in every case (Volume of base + Volume of acid), therefore the molarity will change as well

formulas we will need:

M = moles / volume

pH = -log[H+]

pKa = -log(Ka) = -log(1.8*10^-5) = 4.75

Let

CH3COOH be "HA" for simplicity so it will dissociate as follows:

CH3COOH <--> H+ + CH3COO-

HA <--> H+ + A-

Ka = [H+][CH3COO-]/[CH3COOH]

Ka = [H+][A-]/[HA]

now...

a) No base, only weak acid

initially HA -> H+ + A-

Ka = [H+][A-]/[HA]

Assume [H+] = [A-] = x; due to stoichiometry

[HA] = M – x (x accounts for the dissociated acid in equilibrium)

Substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

1.8*10^-5 = x*x/(0.150-x)

x =0.00163

For pH

pH = -log(H+)

pH =-log(0.00163 )

pH in a = 2.08990

b) when Vbas = 25 ml

This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)

Use Henderson-Hassebach equations!

pH = pKa + log (A-/HA)

where;

A- is the conjugate and HA the potonated acid:

initially

mmol of acid = MacidVacid = 50*0.150 = 7.5 mmol of acid

mmol of base = MbaseVbase = 25*0.150 = 3.75 mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 7.5 - 3.75 = 3.75 mmol

mmol of conjguate left = 0 + 3.75 = 5

Apply equation

pH = pKa + log ([A-]/[HA]) =

pH = 4.75 + log (3.75 /3.75 )

pH = 4.75

c) for 50 ml

Addition of Same quantitie of Acid/Base

This will be Hydrolisis (equilibrium of acid-base) and the weak acid/base will form an equilibrium

We will need Kb

Ka*Kb = Kw

And Kw = 10^-14 always at 25°C for water so:

Kb = Kw/Ka = (10^-14)/(1.8*10^-5) = 5.55*10^-10

Now, proceed to calculate the equilibrium

H2O + A- <-> OH- + HA

Then K equilibrium is given as:

Kb = [HA][OH-]/[A-]

Assume [HA] = [OH-] = x

[A-] = M – x

M = mmol of conjugate / Total V = (7.5) / (50+50) = 0.075 M

Substitute in Kb

5.55*10^-10 = [x^2]/[0.075-x]

x = 6.45*10^-6

[OH-] =6.45*10^-6

Get pOH

pOH = -log(OH-)

pOH = -log (6.45*10^-6) = 5.190

pH = 14-pOH = 14-5.190= 8.81

There will be finally an Excess of Base!

mol of acid < mol of base

Calculate pOH directly

[OH-] = M*V / Vt

mmol of acid = MV = 50*0.150= 7.5

mmol of base = MV = 60*0.150 = 9

therefre,

mmol of strng base left = 9-7.5 = 1.5 mmol

Vtotal = 60+50= 110mL

[OH-] = 1.5 /110 = 0.01363636

pOH = -log(OH-)

pOH = -log(0.01363636) = 1.86530

pH = 14-pOH = 14-1.86530= 12.134

pH = 12.134

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